Solve 10x-9=x^2

$$10 x - 9 = x ^ { 2 }$$

Solve for x

$$x=1$$

$$x=9$$

Steps Using Factoring By Grouping

Subtract x^{2} from both sides.

$$10x-9-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+10x-9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-9. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-9\right)=9$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 9.

$$1,9 3,3$$

Calculate the sum for each pair.

$$1+9=10 3+3=6$$

The solution is the pair that gives sum 10.

$$a=9 b=1$$

Rewrite -x^{2}+10x-9 as \left(-x^{2}+9x\right)+\left(x-9\right).

$$\left(-x^{2}+9x\right)+\left(x-9\right)$$

Factor out -x in -x^{2}+9x.

$$-x\left(x-9\right)+x-9$$

Factor out common term x-9 by using distributive property.

$$\left(x-9\right)\left(-x+1\right)$$

To find equation solutions, solve x-9=0 and -x+1=0.

$$x=9 x=1$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$10x-9-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+10x-9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 10 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-1\right)\left(-9\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-10±\sqrt{100+4\left(-9\right)}}{2\left(-1\right)}$$

Multiply 4 times -9.

$$x=\frac{-10±\sqrt{100-36}}{2\left(-1\right)}$$

Add 100 to -36.

$$x=\frac{-10±\sqrt{64}}{2\left(-1\right)}$$

Take the square root of 64.

$$x=\frac{-10±8}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-10±8}{-2}$$

Now solve the equation x=\frac{-10±8}{-2} when ± is plus. Add -10 to 8.

$$x=-\frac{2}{-2}$$

Divide -2 by -2.

$$x=1$$

Now solve the equation x=\frac{-10±8}{-2} when ± is minus. Subtract 8 from -10.

$$x=-\frac{18}{-2}$$

Divide -18 by -2.

$$x=9$$

The equation is now solved.

$$x=1 x=9$$

Steps for Completing the Square

Subtract x^{2} from both sides.

$$10x-9-x^{2}=0$$

Add 9 to both sides. Anything plus zero gives itself.

$$10x-x^{2}=9$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+10x=9$$

Divide both sides by -1.

$$\frac{-x^{2}+10x}{-1}=\frac{9}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{10}{-1}x=\frac{9}{-1}$$

Divide 10 by -1.

$$x^{2}-10x=\frac{9}{-1}$$

Divide 9 by -1.

$$x^{2}-10x=-9$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-9+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-9+25$$

Add -9 to 25.

$$x^{2}-10x+25=16$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=16$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{16}$$

Simplify.

$$x-5=4 x-5=-4$$

Add 5 to both sides of the equation.

$$x=9 x=1$$